, 2026

The top-left square is divided into $4$ regions having equal area. Since $1$ of these $4$ regions is shaded, then $25\%$ of the square is shaded.

The top-middle square is divided into $8$ regions having equal area. Since $2$ of these $8$ regions are shaded, then $25\%$ of the square is shaded.

The top-right square is divided into $2$ larger regions having equal areas, and
$2$ smaller regions having equal
areas. Since $1$ of the larger
regions is shaded, then more than $25\%$ of the square is shaded.
Alternatively, the shaded region consists of three complete $1\times1$ squares together with $6$ half $1\times1$ squares, and thus the shaded
area is $6$. The entire $4\times4$ square has area $16$, and $\frac{6}{16}$ is not equivalent to $25\%$.

The bottom-left square is divided into $16$ regions having equal area. Since $5$ of these $16$ regions are shaded, then more than
$25\%$ of the square is shaded.

The bottom-middle square is divided into $4$ larger regions having equal areas, and
$4$ smaller regions having equal
areas. Since $2$ of the $4$ larger regions are shaded, and $2$ of the $4$ smaller regions are shaded, then $50\%$ of the square is shaded.

In the bottom-right square, consider constructing a vertical line
segment between the midpoint of the top side of the square and the
midpoint of the bottom side of the square. The square would then be
divided into $4$ regions having equal
areas. Since $1$ of these regions is
shaded, then $25\%$ of the square is
shaded.

Thus, the top-left, top-middle, and bottom-right squares have exactly
$25\%$ of their area shaded and so
there are $3$ such squares.